Proof of limit by epsilon-delta method: $\displaystyle\lim_{x \to 9} \sqrt{x-5} = 2$

Question

The question is:

Prove the below limit statement:

$\displaystyle\lim_{x \to 9} \sqrt{x-5} = 2$

From my understanding of the textbook (Thomas' Calculus), the proof is done in 3 steps:

1. Write both the $\epsilon$ and $\delta$ inequalities.
2. Simplify the $|f(x)-L|<\epsilon$ inequality to the form $0<|x-c|<k\epsilon$, where k is a real number. Set $\delta$ in terms of $\epsilon$
3. Substitute the new $\delta$ inequality in $|f(x)-L|$ to get back the original $\epsilon$ inequality.

(I can include an example of this method from the textbook ($\lim_{x \to 1} \left(5x-3\right)=2$) to demonstrate this method if necessary.)

I tried to follow the same steps with the above mentioned question, but I am unable to perform step 4 as $\delta$ is a minimum of two deltas. Below is my working for reference:

Step 1: $0<|x-9|<\delta$ and $|f(x)-2|<\epsilon$, where $f(x)=\sqrt{x-5}$

Step 2:

$\Rightarrow |\sqrt{x-5}-2|<\epsilon$

$\Rightarrow -\epsilon+2 < \sqrt{x-5} < \epsilon+2$ (adding $-2$ and removing modulus)

$\Rightarrow \left(2-\epsilon\right)^{2} < x-5 < \left(2+\epsilon\right)^{2}$

$\Rightarrow \left(2-\epsilon\right)^{2} + 5 < x < \left(2+\epsilon\right)^{2} + 5$ [1]

Now, the delta inequality states $0<|x-9|<\delta\ \Rightarrow -\delta < x-9 < \delta$

$\Rightarrow -\delta+9 < x < \delta+9$ [2]

Equating [1] and [2],

LHS	RHS
$\Rightarrow -\delta+9 = \left(2-\epsilon\right)^{2} + 5$	$\Rightarrow \delta+9 = \left(2+\epsilon\right)^{2} + 5$
$\Rightarrow -\delta+9 = 4+\epsilon^2-4\epsilon+5$	$\Rightarrow \delta+9 = 4+\epsilon^2+4\epsilon+5$
$\Rightarrow -\delta = \epsilon^2 - 4\epsilon$	$\Rightarrow \delta = \epsilon^2 + 4\epsilon$
$\Rightarrow \delta = 4\epsilon - \epsilon^2$	$\Rightarrow \delta = 4\epsilon + \epsilon^2$

Hence, $\delta = \min\left(4\epsilon - \epsilon^2, 4\epsilon + \epsilon^2\right) = 4\epsilon - \epsilon^2$

Now, proceeding to step 4 yields weird results:

Now $0<|x-9|<(4\epsilon - \epsilon^2)$

$-4\epsilon + \epsilon^2<x-9<(4\epsilon - \epsilon^2)$

$-4\epsilon + \epsilon^2 + 9<x<4\epsilon - \epsilon^2 + 9$

$\epsilon^2 - 4\epsilon + 4 < x-5 < -\epsilon^2 + 4\epsilon + 4$

On completing the square and taking square root...

$\epsilon-2<\sqrt{x-5}<\sqrt{8-\left(\epsilon-2\right)^2}$.

I want to reach back to this inequality: $|\sqrt{x-5} - 2|<\epsilon$.

What am I doing wrong here? Is this step 4 even necessary? (Step 4 is not done in many questions across MSE and other sites, however all examples in Thomas' Calculus perform it.)

Answer 1

You should have assumed $\epsilon<2$ (in step 2). So shall we.

In your (anyway useless) verification, don't forget how you chose $\delta:$ $$\delta=\min\left(4\epsilon - \epsilon^2, 4\epsilon + \epsilon^2\right) = 4\epsilon - \epsilon^2,$$ where $ 4\epsilon + \epsilon^2=(2+\epsilon)^2+5-9$ came from the RHS upperbound in [1]. Conversely: $$\begin{align}|x-9|<4\epsilon-\epsilon^2&\iff\epsilon^2 - 4\epsilon + 4 < x-5 < -\epsilon^2 + 4\epsilon + 4\\&\implies\epsilon^2 - 4\epsilon + 4 < x-5 <\color{red}+\epsilon^2 + 4\epsilon + 4\\ &\iff\dots\\ &\iff|\sqrt{x-5}-2|<\epsilon.\end{align}$$

Answer 2

You made the mistake of trying to solve an equation telling you what delta should be, and you found two equations with different results.

What you need is an upper bound for delta: In your case, delta must be at most 4 epsilon minus epsilon squared, and it must be at most 4 epsilon plus epsilon squared. And that’s easy: You pick delta less than the smaller number.